Is MIT Prof. Lewin wrong about Kirchhoff's law?

[yungman]

Please provide a diagram. My xfmr example demonstrated that KVL does not always work. We are not in agreement here.

Claude

We never in agreement, I just use your example to build on it. I scan the drawing, don't know how long before people can see it, I know there will be a delay. If you have any easier way to post the picture, tell me how.

I label the points exactly the same as the example I gave on the single turn loop of the professor. The wire is the secondary of the transformer.

BTW, I agree the inductance is not in the picture just like you said your secondary internal impedance is 1ohm. I posted the calculations on the inductance of the 24 gauge wire taking into consideration of skin effect and all. Turn out that it is only a few ohms and not significant in the whole picture.

 

[yungman]

I & others have stated such. You are making much ado about nothing. Including a voltage source is nothing but a construct. The "voltage source" (or current source per Norton/Thevenin equivalence principle) added to the circuit makes KVL valid. What does that mean? It means that w/o said voltage source, KVL is invalid. Dr. Lewin made this point, a correct point at that.

In reality, there is no "voltage source" (nor "current source) in series/parallel w/ the loop. The induced emf/mmf is distributed around the loop. The net voltage around the loop is not zero, but rather, the induced emf. KVL does not hold. Adding the voltage source to the loop modifies the problem by replacing distributed quantities w/ lumped quantities. Then KVL holds because we've transformed the problem from fields to circuits.

Dr. Lewin stated all this, & he has it right. His critics think they know more than him & other learned people. They don't. The problem with the critics is that they don't know what they don't know. They make much ado about things that are very well known. "Is Prof. Lewin wrong about Kirchoff's law?" is the title of this thread.

No, he is not wrong. He is right.

Claude

How can you not include the voltage source into the circuit. Without that, you cannot even generate the current to give the voltage across the two resistors. I guess you understand what I am driving at that there should be a voltage source in the drawing. I cannot see how you can not consider that is part of the circuit. KVL do use voltage source.

Induce EMF is a voltage and it is a source. It might be distribute over the whole loop but if you draw the equivalent circuit as all the books do, you replace the distribute source as a single source. Look at the equivalent schematic of Transmission line that represent the distribute L, R, G and C, people use discrete resistor, inductor etc. to represent the distributed element per unit length. How can you not consider the distribute emf on the loop as not a voltage source?

The professor was wrong. Until we can get over the way he described on his initial drawing, everything else is irrelevant.

Let me put it this way, ALL voltage source are generated, if you look at ALL the examples in books of KVL with voltage source and SHORT them all out! Everyone of the example failed! Then KVL is hot air if that is your point.

 

[yungman]

I broke down and watched the first video. I found Lewin to be irritating. He seems to get too much delighted in generating confusion rather than clarity.

You can measure the same two physical points and get two different measurements because the leads of the measuring instrument enclose different regions of changing magnetic flux. It's really that simple.

The fact that this crazy thread has gone on so long is evidence of the guy's overwhelming success in creating confusion. Then he gets to be the genius-hero and rescue you from the confusion he, himself has so cleverly led you into. good grief.

Clear the indoctrination of this subversive screwball out of your head, learn about electromagnetic fields, then come back to it, and the confusion will have evaporated.

I don't know that guy, I thought his presentation is just simply wrong. I am absolute amazed that people don't see the problem of his presentation and keep on arguing on the formulas behind the non conservative nature this and that. To me, until he make his model correctly, there is no merits on the rest.

I found him arrogant and condescending. Apparently people here respect him and his big name. I worked over 10 years in a company with over 60% PHDs as staff, this is nothing new. I spent years arguing with them! They are just human and they make mistake. And I notice a lot of them have such ego that they defend to death! We are no MIT, but you should see the draws of publications in scientific journals from our company. I myself have two paper published in the Review of Scientific Instrument in AIP and own a pattern on the detector of one of the product.

 

[cabraham]

How can you not include the voltage source into the circuit. Without that, you cannot even generate the current to give the voltage across the two resistors. I guess you understand what I am driving at that there should be a voltage source in the drawing. I cannot see how you can not consider that is part of the circuit. KVL do use voltage source.

Induce EMF is a voltage and it is a source. It might be distribute over the whole loop but if you draw the equivalent circuit as all the books do, you replace the distribute source as a single source. Look at the equivalent schematic of Transmission line that represent the distribute L, R, G and C, people use discrete resistor, inductor etc. to represent the distributed element per unit length. How can you not consider the distribute emf on the loop as not a voltage source?

The professor was wrong. Until we can get over the way he described on his initial drawing, everything else is irrelevant.

Let me put it this way, ALL voltage source are generated, if you look at ALL the examples in books of KVL with voltage source and SHORT them all out! Everyone of the example failed! Then KVL is hot air if that is your point.

How can we not include the voltage source is a good question. Dr. Lewin is merely illustrating that the definition of voltage across 2 points depends on the path chosen. Since integral E*dl is voltage, the path around the loop is the resistors & the conductors. There is no physical voltage source, or current source, present in the circuit.

We can, however, model an equivalent circuit which includes a voltage/current source (Thevenin/Norton). Then, KVL will be upheld. Without a source included in the circuit, KVL does not always hold.

It's that simple. In your sketch you attached, you are including a source, namely the xfmr secondary. When summing voltages around the secondary loop, the result is zero, regardless of path. In other words, KVL is valid here. But you have lumped the quantities. That in & of itself is not wrong, it just gives a different result.

If we drew a loop consisting of 2 resistors connected by wires in a closed loop, & do not lump the induction into a discrete source, then KVL does not hold. The sum of voltages around the loop equals the induced emf. But if we insert a lumped voltage source into the loop, whose value equals the induced emf, then the induced emf is canceled by the source, leaving zero around the loop. Thus KVL holds in this condition.

There seems to be universal consensus here that if the discrete lumped source equalling the induced emf is added to the circuit, KVL is upheld. If not, then KVL may not apply. I think that sums it up.

No need to argue. I'll gladly clarify further if needed. Again, I don't share Dr. Lewin's method of presenting his facts. His facts are correct, but I believe I could explain said facts in a manner which freshman & sophomores could relate to. Then again, I may be mistaken. I understand the material well. But the ability to convey it to a novice may not be easy. I might be overestimating my ability to convey info. It happens.

Claude

 

[hikaru1221]

@yungman (& some other people): From what I understand, your transformer effect is actually included in Prof. Lewin's explanation. It's nothing else but electromagnetic induction. I don't understand where your transformer effect violates with Prof. Lewin's explanation. The "transformer" here is simply a very-non-ideal transformer, and thus, using the term "induction" is more appropriate.
Besides, since the wire's inductance is negligible, the 1-volt emf is mostly distributed on the 2 resistors (not evenly distributed over the whole loop). I don't really understand what to be in dispute here.

P.S.: I've just had a look at the file you attached in post #106. I think we cannot treat the circuit this way. The inductor (one of the coils of the transformer in particular) is MODELED to fit in simplified version of KVL; the emf is on the inductor. The circuit of Prof. Lewin is not that way; the emf is on the resistors.

P.S. #2: I think there is one thing that should be clarified. The induced emf due to the coil is NOT the only emf here; in other words, induced E-field is not the only E-field here. There is also E-field due to charge accumulation at the boundary between the resistors and the wires (since they are 2 different mediums with different resistivity!).
The induced E-field exists in the space regardless of the presence of the circuit; therefore, there is induced emf on the wire. However, the E-field due to charges cancels it. The final result is that the wire has zero emf on it, the 2 resistors have 1-volt emf.@Studiot: I think there is some other semantics here. From what I understand, your "Faraday's law" is the statement / the theory proposed by Faraday. The Faraday's law, as far as I've known, is the law that states induced emf is proportional to rate of change in magnetic flux (so "Faraday's law" is just a name). According to Wikipedia, in an attempt to explain the phenomenon, Faraday did propose a notion of line of force, which was rejected. Faraday's idea may come farther than the law, but the law which is named after Faraday has nothing to do with differential form or the notion of induced E-field.

 

[sarumonkee]

How can we not include the voltage source is a good question. Dr. Lewin is merely illustrating that the definition of voltage across 2 points depends on the path chosen. Since integral E*dl is voltage, the path around the loop is the resistors & the conductors. There is no physical voltage source, or current source, present in the circuit.

We can, however, model an equivalent circuit which includes a voltage/current source (Thevenin/Norton). Then, KVL will be upheld. Without a source included in the circuit, KVL does not always hold.

It's that simple. In your sketch you attached, you are including a source, namely the xfmr secondary. When summing voltages around the secondary loop, the result is zero, regardless of path. In other words, KVL is valid here. But you have lumped the quantities. That in & of itself is not wrong, it just gives a different result.

If we drew a loop consisting of 2 resistors connected by wires in a closed loop, & do not lump the induction into a discrete source, then KVL does not hold. The sum of voltages around the loop equals the induced emf. But if we insert a lumped voltage source into the loop, whose value equals the induced emf, then the induced emf is canceled by the source, leaving zero around the loop. Thus KVL holds in this condition.

There seems to be universal consensus here that if the discrete lumped source equalling the induced emf is added to the circuit, KVL is upheld. If not, then KVL may not apply. I think that sums it up.

No need to argue. I'll gladly clarify further if needed. Again, I don't share Dr. Lewin's method of presenting his facts. His facts are correct, but I believe I could explain said facts in a manner which freshman & sophomores could relate to. Then again, I may be mistaken. I understand the material well. But the ability to convey it to a novice may not be easy. I might be overestimating my ability to convey info. It happens.

Claude

I don't get your comment. You say if the model is changed to include an element that affects the circuit, you get the right answer. If the model ignores the "lumped element" that is the voltage source (or inductor), then it is not wrong? How can you justify ignoring the effect of the very thing that is coupling energy into the system? I just don't get it. Maybe it is a difference in thought between physicists and engineers...

Just because something is "spread out" around a circuit doesn't mean we can ignore it and remove it from the model. It seems like that is akin to saying a resistor can be ignored because it is a collection of atoms, and therefore the resistance is distributed.

 

[sarumonkee]

@yungman (& some other people): From what I understand, your transformer effect is actually included in Prof. Lewin's explanation. It's nothing else but electromagnetic induction. I don't understand where your transformer effect violates with Prof. Lewin's explanation. The "transformer" here is simply a very-non-ideal transformer, and thus, using the term "induction" is more appropriate.
Besides, since the wire's inductance is negligible, the 1-volt emf is mostly distributed on the 2 resistors (not evenly distributed over the whole loop). I don't really understand what to be in dispute here.

The wire's inductance is not negligible... It is what allows the energy to be coupled into the circuit. That's why Lewin has such a huge coil and says it "BLASTS" flux everywhere. Yeah, the inductance is small, but with enough energy, Lewin is coupling the needed energy to get the currents he does through the resistors.

 

[hikaru1221]

The wire's inductance is not negligible... It is what allows the energy to be coupled into the circuit. That's why Lewin has such a huge coil and says it "BLASTS" flux everywhere. Yeah, the inductance is small, but with enough energy, Lewin is coupling the needed energy to get the currents he does through the resistors.

What I meant by "wire" is the connecting wire of the circuit, the "secondary coil" of the transformer, not the wire of the primary one. Sorry for the confusion.

 

[atyy]

The wire's inductance is not negligible... It is what allows the energy to be coupled into the circuit. That's why Lewin has such a huge coil and says it "BLASTS" flux everywhere. Yeah, the inductance is small, but with enough energy, Lewin is coupling the needed energy to get the currents he does through the resistors.

No one is saying the wire's inductance is negligible.

It is the whole thing.

The question is how do you model the inductance?

By a kluge lumped element, or by Faraday's law?

A lumped element is fine for many purposes, but if you want to predict the different readings of voltmeters connected to exactly the same points, then you need Faraday's law.

 

[yungman]

How can we not include the voltage source is a good question. Dr. Lewin is merely illustrating that the definition of voltage across 2 points depends on the path chosen. Since integral E*dl is voltage, the path around the loop is the resistors & the conductors. There is no physical voltage source, or current source, present in the circuit.

We can, however, model an equivalent circuit which includes a voltage/current source (Thevenin/Norton). Then, KVL will be upheld. Without a source included in the circuit, KVL does not always hold.

It's that simple. In your sketch you attached, you are including a source, namely the xfmr secondary. When summing voltages around the secondary loop, the result is zero, regardless of path. In other words, KVL is valid here. But you have lumped the quantities. That in & of itself is not wrong, it just gives a different result.

If we drew a loop consisting of 2 resistors connected by wires in a closed loop, & do not lump the induction into a discrete source, then KVL does not hold. The sum of voltages around the loop equals the induced emf. But if we insert a lumped voltage source into the loop, whose value equals the induced emf, then the induced emf is canceled by the source, leaving zero around the loop. Thus KVL holds in this condition.
But in real life, the 6" wire is the secondary of the transformer. Yes the resistor body is part of the loop that pickup the magnetic field and generate part of the emf. But common sense is that a 1/4W resistor is about 0.3" length and two including the point to point solder is about 0.7" which is about 10% of the total length of the loop. So the error is 10% of the 9:1 ratio. This is within the error of the scope measurement easily. I just ignor the resistor body effect like you ignor the 1ohm internal impedance ( which I agree) of the secondary winding. We are arguing about the 9:1 ratio, not the 10% small stuff.
There seems to be universal consensus here that if the discrete lumped source equalling the induced emf is added to the circuit, KVL is upheld. If not, then KVL may not apply. I think that sums it up.

No need to argue. I'll gladly clarify further if needed. Again, I don't share Dr. Lewin's method of presenting his facts. His facts are correct, but I believe I could explain said facts in a manner which freshman & sophomores could relate to. Then again, I may be mistaken. I understand the material well. But the ability to convey it to a novice may not be easy. I might be overestimating my ability to convey info. It happens.

Claude

I consider this is a friendly debate here, no hard feeling at all. I just believe we have to include the voltage source no mater what, without the voltage source, where is the current come from? No voltage source, there is nothing for us to talk about. I model this as a single voltage source, this is in line with every electronics book that I read. Just use the model of the transmission that I described, you really don't have individual resistor, inductor etc. It is just a distributed element and is expressed as ohm/length, farad/length or henry/length. This transmission line model is in EM or ED textbook too!

 

[yungman]

No one is saying the wire's inductance is negligible.

It is the whole thing.

The question is how do you model the inductance?

By a kluge lumped element, or by Faraday's law?

A lumped element is fine for many purposes, but if you want to predict the different readings of voltmeters connected to exactly the same points, then you need Faraday's law.

Remember Sarumonkee repeat the experiment in #90 and he showed if he reference the ground of the probe in the middle of the wire ( 3” from each side) and measure the junction of the wire and resistor on both side, he got equal and opposite voltage on the wire. So we know FOR FACT there is induced emf on the wire and it is the emf that drive the resistors and get the 9:1 ratio of voltage.

 

[yungman]

@yungman (& some other people): From what I understand, your transformer effect is actually included in Prof. Lewin's explanation. It's nothing else but electromagnetic induction. I don't understand where your transformer effect violates with Prof. Lewin's explanation. The "transformer" here is simply a very-non-ideal transformer, and thus, using the term "induction" is more appropriate.
Besides, since the wire's inductance is negligible, the 1-volt emf is mostly distributed on the 2 resistors (not evenly distributed over the whole loop). I don't really understand what to be in dispute here.
Not he did not include the transformer in his drawing, he just use point A and D between the two resistors. Double check part one of the video. That was when he start flying off the handle on trashing others. If there is not source, where is the 1mA that drive the resistors come from. You cannot ignor the source. If you look at every example of KVL and eliminate all the voltage source that don't conform to the definition of descrete element, KVL fail most of the time!
P.S.: I've just had a look at the file you attached in post #106. I think we cannot treat the circuit this way. The inductor (one of the coils of the transformer in particular) is MODELED to fit in simplified version of KVL; the emf is on the inductor. The circuit of Prof. Lewin is not that way; the emf is on the resistors.
Look at my post #115 in blue color. I gave the detail explanation about the resistor. The resistor only contribute about 10% of the induction, the wire generate the body of the emf.
P.S. #2: I think there is one thing that should be clarified. The induced emf due to the coil is NOT the only emf here; in other words, induced E-field is not the only E-field here. There is also E-field due to charge accumulation at the boundary between the resistors and the wires (since they are 2 different mediums with different resistivity!).
The junction potential are small, we are talking about the dispute of 9:1, I just simply ignor anything that is under 10%. The measurement by the professor is only accurate to about 10% in my estimate.

The induced E-field exists in the space regardless of the presence of the circuit; therefore, there is induced emf on the wire. However, the E-field due to charges cancels it. The final result is that the wire has zero emf on it, the 2 resistors have 1-volt emf.


@Studiot: I think there is some other semantics here. From what I understand, your "Faraday's law" is the statement / the theory proposed by Faraday. The Faraday's law, as far as I've known, is the law that states induced emf is proportional to rate of change in magnetic flux (so "Faraday's law" is just a name). According to Wikipedia, in an attempt to explain the phenomenon, Faraday did propose a notion of line of force, which was rejected. Faraday's idea may come farther than the law, but the law which is named after Faraday has nothing to do with differential form or the notion of induced E-field.

As I said, context is everything here.

 

[yungman]

I want to start this new post to remind everyone to read #90 from Samumonkee that he repeated the experiment and measure the voltage across the wire.

I really have to get back to my study, I'll try to resist checking this thread until I get some of my own work done!

 

[stevenb]

So we know FOR FACT there is induced emf on the wire and it is the emf that drive the resistors and get the 9:1 ratio of voltage.

Why do you think that emf on the wire is a surprise to us? Faraday's law tells us that there is a 1V emf around the loop. This new setup of sarumonkee has tiny leads on the resistor that are soldered right next to each other, and then a long length of wire to finish the loop. Wouldn't we expect emf in the wire?

Faraday's law tells us something else about this new measurement setup. There is also emf on the scope wires. Why? Because the scope leads are a valid completion of the loop with equal validity as the main circuit wire. For this reason, if you take a loop through your scope and probes and complete it through the wire, the net emf is zero because the wire and the scope leads have equal emf in opposite directions around this new loop. This makes sense in terms of Faraday's Law because there is no net flux change inside this new loop.

What is mysterious to me is that sarumonkee got a reading of emf on the scope when the loop emf being measured should be zero. Yes, there can be emf on the wire, but not in the measurement loop. This is the underlying reason that Lewin's analysis is correct. When you analyze a system with Faraday's law using loops, these details take care of themselves, and you don't need to identify the location of the emf. So, I'm highly suspect of the measurements of sarumonkee, but it's quite possible I'm misinterpreting the method he used. Unless I see the layout and know the method, I can't really judge. I am waiting for him to do more verifications and checking and then to post the details so we can be convinced of his methods. He should follow Lewin's lead and verify all loops obey Faraday's Law. This is a good check to help give confidence that the measurement techniques are valid. I commend sarumonkey for doing real measurements and getting to the bottom of this personally, and it's only fair to give him adequate time and not rush him.

 

[sarumonkee]

No one is saying the wire's inductance is negligible.

It is the whole thing.

The question is how do you model the inductance?

By a kluge lumped element, or by Faraday's law?

A lumped element is fine for many purposes, but if you want to predict the different readings of voltmeters connected to exactly the same points, then you need Faraday's law.

Sorry I didn't quote properly. That was a response to Hikaru's post #110 where s/he said:

"Besides, since the wire's inductance is negligible, the 1-volt emf is mostly distributed on the 2 resistors (not evenly distributed over the whole loop). I don't really understand what to be in dispute here."

Also, I didn't put the probes on the exact same spot yet, so I can't comment on that yet. I probably won't have time to do the testing again until this weekend, if that...

 

[cabraham]

I don't get your comment. You say if the model is changed to include an element that affects the circuit, you get the right answer. If the model ignores the "lumped element" that is the voltage source (or inductor), then it is not wrong? How can you justify ignoring the effect of the very thing that is coupling energy into the system? I just don't get it. Maybe it is a difference in thought between physicists and engineers...

Just because something is "spread out" around a circuit doesn't mean we can ignore it and remove it from the model. It seems like that is akin to saying a resistor can be ignored because it is a collection of atoms, and therefore the resistance is distributed.

Well, let's say you are measuring the circuit, unaware that induction is taking place. You measure the voltage along different paths, & the sum around the loop is non-zero. This should tell you that induction is going on. Of course, the mag field is what couples energy into the loop. But if you don't see it, & are unaware of its presence, you will make measurements at odds with KVL. That is the point.

Dr. Lewin was merely illustrating that the sum of voltages around a loop may be non-zero. If non-zero, however, the value is exactly equal to the induced emf. Thus if said emf is added to the circuit model in the form of an independent source/generator, whose voltage value equals that of the induced emf, then it will balance & KVL will hold, as the sum around the loop is now zero.

What I take from all this is that fields are distributed parameters, & circuits are lumped. When jumping between the two, we must be careful. In circuits, the sum around a loop is zero per KVL. If the circuit itself is acting as an inductance with an incident time varying mag field upon it, then the sum around the loop is the induced emf, not zero. Lumping the induced emf into a discrete source balances the loop restoring zero net voltage, & KVL is valid.

Pretty easy, if you ask me.

Claude

 

[hikaru1221]

@yungman: The source is connected to the other coil, which propels the magnetic field, so how can it be drawn in a circuit which doesn't contain it? The professor was drawing a schematic diagram of the real set-up, not any model.
Besides, please look at the bold part of my previous post. There is induced emf on the wire, and by your estimation, it constitutes 90% of the total induced emf. However, this is not the only emf inside the circuit. Due to this extra emf, the 90% is simply "shifted" from the wire to the resistors. The equation for a loop remains the same, but the real thing happening inside the circuit is all that is on the resistors. The extra emf is NOT LESS THAN 10%.
Thanks for joining the debate and all the best to your work

@sarumonkee: I would like to see your experiment with my own eyes and learn, too

 

[yungman]

Why do you think that emf on the wire is a surprise to us? Faraday's law tells us that there is a 1V emf around the loop. This new setup of sarumonkee has tiny leads on the resistor that are soldered right next to each other, and then a long length of wire to finish the loop. Wouldn't we expect emf in the wire?

Faraday's law tells us something else about this new measurement setup. There is also emf on the scope wires. Why? Because the scope leads are a valid completion of the loop with equal validity as the main circuit wire. For this reason, if you take a loop through your scope and probes and complete it through the wire, the net emf is zero because the wire and the scope leads have equal emf in opposite directions around this new loop. This makes sense in terms of Faraday's Law because there is no net flux change inside this new loop.

What is mysterious to me is that sarumonkee got a reading of emf on the scope when the loop emf being measured should be zero. Yes, there can be emf on the wire, but not in the measurement loop. This is the underlying reason that Lewin's analysis is correct. When you analyze a system with Faraday's law using loops, these details take care of themselves, and you don't need to identify the location of the emf. So, I'm highly suspect of the measurements of sarumonkee, but it's quite possible I'm misinterpreting the method he used. Unless I see the layout and know the method, I can't really judge. I am waiting for him to do more verifications and checking and then to post the details so we can be convinced of his methods. He should follow Lewin's lead and verify all loops obey Faraday's Law. This is a good check to help give confidence that the measurement techniques are valid. I commend sarumonkey for doing real measurements and getting to the bottom of this personally, and it's only fair to give him adequate time and not rush him.

OK, my bad, I can't resist to take a peek here even though I should take more peek at my books.

Sarumonkee did the wise thing of waving the probe around and did not see any changes. That tells a lot. Say the setup of the resistor is on the xy plane, the B is on z axis, you wave the probe around, the loop of the probe and the wire can be tangent to the B or on xy plane but out side the path of the B. If the magnetic field is so strong, he would have seen big changes on the reading.

Anyway, I drawn up a setup to minimize the loop area of the measurement and I attach here. I use two coax, solder the ground shield at the middle point of the wire loop as shown. the cable inside the coax should twist with the wire of the loop to ensure the area between the wire and the coax inner cable is kept to minimum. The ability of picking up mag field is proportional to the area inside the loop, by minimize the area, you minimize the pickup. This will give a better reading. If you want to. twist the two coax together to minimize the area also until ending to the scope farther away.

I am gone, I'll try harder to stay away until tonight!

 

[cabraham]

I consider this is a friendly debate here, no hard feeling at all. I just believe we have to include the voltage source no mater what, without the voltage source, where is the current come from? No voltage source, there is nothing for us to talk about. I model this as a single voltage source, this is in line with every electronics book that I read. Just use the model of the transmission that I described, you really don't have individual resistor, inductor etc. It is just a distributed element and is expressed as ohm/length, farad/length or henry/length. This transmission line model is in EM or ED textbook too!

The current is coming from the time varying E & H fields. That's where. The Lorentz force equation is:

F = q*(E + u X B).

The reason charges move through the loop is Lorentz force, which can be expressed in terms of the fields & velocity. Regarding the inclusion or lack of a discrete source, here goes.

In distributed field terms, w/o source --> Vr1 + Vr2 = Vinduced, for a loop w/ 2 resistors.

In lumped circuit terms, w/ source --> Vr1 +Vr2 - Vinduced = 0, KVL applies.

There is no difference at all. You do not have to include a discrete source. Everything is included. Without a source, the summation of loop voltage equals the induced emf. We are acknowledging said induced emf as the sum around the loop. Without the source included, what is driving the current? Answer: the time varying fields carry energy/power. The power induced cannot exceed the watts in the incident field.

With lumped parameter equivalent circuit including a source, the sum of voltage drops equals the induced emf. But the equation is written w/ the induced emf on the same side as the resistive voltage drops, resulting in a sum of zero, which agrees with KVL. But the 2 eqns above are identical, because if a + b = c, then it also holds that a + b - c = 0, as well. The 2nd form of the equation looks like KVL. But both are equivalent.

Include the source, or don't include it. Lump the quantities, or distribute them. As long as you account for everything you should get the right answer either way. Again, Dr. Lewin's critics are making too much ado over nothing.

I designed magnetics for a living full time, then branched into broader electronic r&d while still doing magnetics as well, for 32 yrs. now. I'm in the final stage of the Ph.D. program. I had to study e/m fields for the qualifier. This stuff is not trivial, & can be confusing. But great minds have figured it out. The critics are blowing smoke, as they don't know as much as they think. Cheers.

Claude

 

[yungman]

The current is coming from the time varying E & H fields. That's where. The Lorentz force equation is:

F = q*(E + u X B).

The reason charges move through the loop is Lorentz force, which can be expressed in terms of the fields & velocity. Regarding the inclusion or lack of a discrete source, here goes.

In distributed field terms, w/o source --> Vr1 + Vr2 = Vinduced, for a loop w/ 2 resistors.

In lumped circuit terms, w/ source --> Vr1 +Vr2 - Vinduced = 0, KVL applies.
This two is the same equation, Vinduced is there.
There is no difference at all. You do not have to include a discrete source. Everything is included. Without a source, the summation of loop voltage equals the induced emf. We are acknowledging said induced emf as the sum around the loop. Without the source included, what is driving the current? Answer: the time varying fields carry energy/power. The power induced cannot exceed the watts in the incident field.
As I said over and over, the point D is NOT a node! I am waiting for Sarumonkee to repeat the experiment.

With lumped parameter equivalent circuit including a source, the sum of voltage drops equals the induced emf. But the equation is written w/ the induced emf on the same side as the resistive voltage drops, resulting in a sum of zero, which agrees with KVL. But the 2 eqns above are identical, because if a + b = c, then it also holds that a + b - c = 0, as well. The 2nd form of the equation looks like KVL. But both are equivalent.

Include the source, or don't include it. Lump the quantities, or distribute them. As long as you account for everything you should get the right answer either way. Again, Dr. Lewin's critics are making too much ado over nothing.

I designed magnetics for a living full time, then branched into broader electronic r&d while still doing magnetics as well, for 32 yrs. now. I'm in the final stage of the Ph.D. program. I had to study e/m fields for the qualifier. This stuff is not trivial, & can be confusing. But great minds have figured it out. The critics are blowing smoke, as they don't know as much as they think. Cheers.

Claude

I don't think we can see eye to eye, I am talking about the experiment that the professor did and the drawing he had. My only question to you at this point is going back to my first schematic. If you say the source need not be there and it does not matter. So point B and point C is the same point. Are you telling me voltage from point B to point A IS the same as from point C to point A. In all your argument that their are no source, that you should get the same measurement.

$$\hbox { If } V_{BA}=V_{CA}$$

If so, then I am blowing off hot air in this thread. If it is not, then you are going to have to explain this out. I am not working and I don't have all the components readily for me to try it easily. Sarumonkee did the experiment and found voltage across the wire, seems like you just chuck it into his measurement error. Until you can prove that there is no voltage drop across the wire and point B and C are identical electrically, all your Lorenze and FL means nothing. Generating current with any reason still a source and need to be accounted for. I am studying EM and RF for the last 10 years also after 27 years of being a senior engineer and manager of EE. You cannot just look at the equation, setup is everything. You don't know the professor's setup, you can't comment on that. From the drawing, I know point B is not equal to point C.

Being said that, I was thinking, if the loop is consisted only of resistor material, say 900ohm from $0< \phi< \pi$ and 100ohm from $\pi< \phi< 2\pi$, and measure the voltage across point A and D like the professor. I wonder what would be the voltage. I have a suspicion that the voltage is not going to be 9:1. I think on first pass, the voltage is the super-position of the induced emf of the loop and the voltage divider effect. I don't know, that would be interesting if there is resistor like this, a half circle shape. But I am sure what ever the voltage measured, it is path independent. The closest is using those high voltage resistor from brand like Caddock that is 2" long and 1/8" diameter and solder end to end to form a square loop. 3 resistor of say 50ohm and one 900ohm to get about 9:1 ratio ( not very important as long as it is taken into consideration). Then you can measure the voltage across each resistor. Now you almost get 4 notes and you can really try KVL clockwise or CCW.

 

[Studiot]

Dr. Lewin was merely illustrating that the sum of voltages around a loop may be non-zero. If non-zero, however, the value is exactly equal to the induced emf. Thus if said emf is added to the circuit model in the form of an independent source/generator, whose voltage value equals that of the induced emf, then it will balance & KVL will hold, as the sum around the loop is now zero.

What I take from all this is that fields are distributed parameters, & circuits are lumped. When jumping between the two, we must be careful. In circuits, the sum around a loop is zero per KVL. If the circuit itself is acting as an inductance with an incident time varying mag field upon it, then the sum around the loop is the induced emf, not zero. Lumping the induced emf into a discrete source balances the loop restoring zero net voltage, & KVL is valid.

I would like to inject a note of caution here.

If you do this and the loop is part of a larger network or mesh you may again obtain misleading results.

It was for this reason, until the rise of matrix theory for asembling large systems of linear equations for computers, that KVL was originally stated, and taught at MIT and other learned institutions, in the form I presented.

 

[cabraham]

I don't think we can see eye to eye, I am talking about the experiment that the professor did and the drawing he had. My only question to you at this point is going back to my first schematic. If you say the source need not be there and it does not matter. So point B and point C is the same point. Are you telling me voltage from point B to point A IS the same as from point C to point A. In all your argument that their are no source, that you should get the same measurement.

$$\hbox { If } V_{BA}=V_{CA}$$

If so, then I am blowing off hot air in this thread. If it is not, then you are going to have to explain this out. I am not working and I don't have all the components readily for me to try it easily. Sarumonkee did the experiment and found voltage across the wire, seems like you just chuck it into his measurement error. Until you can prove that there is no voltage drop across the wire and point B and C are identical electrically, all your Lorenze and FL means nothing. Generating current with any reason still a source and need to be accounted for. I am studying EM and RF for the last 10 years also after 27 years of being a senior engineer and manager of EE. You cannot just look at the equation, setup is everything. You don't know the professor's setup, you can't comment on that. From the drawing, I know point B is not equal to point C.

Being said that, I was thinking, if the loop is consisted only of resistor material, say 900ohm from $0< \phi< \pi$ and 100ohm from $\pi< \phi< 2\pi$, and measure the voltage across point A and D like the professor. I wonder what would be the voltage. I have a suspicion that the voltage is not going to be 9:1. I think on first pass, the voltage is the super-position of the induced emf of the loop and the voltage divider effect. I don't know, that would be interesting if there is resistor like this, a half circle shape. But I am sure what ever the voltage measured, it is path independent. The closest is using those high voltage resistor from brand like Caddock that is 2" long and 1/8" diameter and solder end to end to form a square loop. 3 resistor of say 50ohm and one 900ohm to get about 9:1 ratio ( not very important as long as it is taken into consideration). Then you can measure the voltage across each resistor. Now you almost get 4 notes and you can really try KVL clockwise or CCW.

In your schematic the discrete source physically exists. It drives the primary of the xfmr, & the resistors are across the secondary. You have to include the source because the induction is confined to the interior of the xfmr. KVL holds here.

The Dr. Lewin setup relies on the distributed nature of induction. When the loop itself forms an inductance, or turn(s) of a xfmr, then KVL must be modified. The summation around said loop is not zero, but the induced emf. If we wish to lump the induction into a discrete source, then KVL will apply.

Studiot mentioned the large network & mesh problem. I am aware of that, but this case involves a simple 1-loop setup. We cannot obtain full agreement with the simple case, so I advise against complicating the issue until we settle this one. I have a multi-loop problem worked out somewhere, if I canot find it I'll recreate & post later. BR.

Claude

 

[sarumonkee]

In your schematic the discrete source physically exists. It drives the primary of the xfmr, & the resistors are across the secondary. You have to include the source because the induction is confined to the interior of the xfmr. KVL holds here.

The Dr. Lewin setup relies on the distributed nature of induction. When the loop itself forms an inductance, or turn(s) of a xfmr, then KVL must be modified. The summation around said loop is not zero, but the induced emf. If we wish to lump the induction into a discrete source, then KVL will apply.

Studiot mentioned the large network & mesh problem. I am aware of that, but this case involves a simple 1-loop setup. We cannot obtain full agreement with the simple case, so I advise against complicating the issue until we settle this one. I have a multi-loop problem worked out somewhere, if I canot find it I'll recreate & post later. BR.

Claude

So after all these pages, and many words, I think I finally realize that we all might be saying the same thing. (Well most of us, some people that didn't stick around past page 4 said some pretty weird things).

Basically, my problem with Lewin's example was he didn't account for the inductance of the wire, making it look like a node had two voltages. From an electrical engineers perspective, that is heresy, as we always attempt to make a model that mimics reality as close as possible, be it lumped element, or what. I am getting the feeling that physicist's sometimes disregard some aspects of a circuit, but I don't have a good sense on why yet.

All the talk about Lewin setting up a strange definition of KVL is making more sense to me, as KVL does not hold, IFF you claim there is no inductor lumped element.

I still contend this makes Lewin look like he doesn't know how energy is being coupled into the circuit, or is being a totally misleading jester...

All in all, I will continue to use the lumped element models for cases like this, as this is most definitely the "most intuitive" thing to do for me :).

 

[stevenb]

From an electrical engineers perspective, that is heresy, as we always attempt to make a model that mimics reality as close as possible,

versus ?

All in all, I will continue to use the lumped element models for cases like this, as this is most definitely the "most intuitive" thing to do for me :).

You have been so critical of Prof. Lewin, but do you even notice your own contradictory statements?

 

[sarumonkee]

versus ?

You have been so critical of Prof. Lewin, but do you even notice your own contradictory statements?

Nope... please enlighten me.

I thought we were under the consensus that there is inductance in the line, Lewin did not have it in his model, and so KVL did work. Are you saying there is no voltage drop across the wires connecting the resistors?

Where are my statements contradictory? I chose to model with lumped elements, instead of completely ignoring their existence... You are confusing me again...

 

[atyy]

So after all these pages, and many words, I think I finally realize that we all might be saying the same thing. (Well most of us, some people that didn't stick around past page 4 said some pretty weird things).

Basically, my problem with Lewin's example was he didn't account for the inductance of the wire, making it look like a node had two voltages. From an electrical engineers perspective, that is heresy, as we always attempt to make a model that mimics reality as close as possible, be it lumped element, or what. I am getting the feeling that physicist's sometimes disregard some aspects of a circuit, but I don't have a good sense on why yet.

All the talk about Lewin setting up a strange definition of KVL is making more sense to me, as KVL does not hold, IFF you claim there is no inductor lumped element.

I still contend this makes Lewin look like he doesn't know how energy is being coupled into the circuit, or is being a totally misleading jester...

All in all, I will continue to use the lumped element models for cases like this, as this is most definitely the "most intuitive" thing to do for me :).

There is an inductance and resistance that Lewin is negelecting by not putting the voltmeters at exactly the same point. Let's call this inductance Lp. It can be modeled as a lumped elements.

There is a completely different inductance due to the big loop of two resistors and wires. Let's call this inductance Lq In Lewin's case Lq is far greater than Lp, so he neglects Lp. He treats Lq using Faraday's law. Some aspects of Lq can be treated by making it a lumped element, but not all. Lq causes the voltmeters to have different readings even if they are connected to exactly the same points.

If you won't believe a physicist, how about two electrical engineers (unfortunately from the same institution) http://web.mit.edu/6.013_book/www/, section 8.4: "In fact, as we now take care to define the circumstances required to make the terminal voltage of a coil a well-defined variable ... If a time-varying magnetic field is significant in this region, then different arrangements of the leads connecting the terminals to the voltmeter will result in different voltmeter readings."

 

[stevenb]

Nope... please enlighten me.

I did enlighten you. Look at the two statements that I quoted from you. You don't see the contradiction? I can enlighten you more if you can't see it.

Lumped models are not models that mimic reality as closely as possible. The fact is that we use approximate models all the time, both in engineering and in phyisics, so your statement that electrical engineers always use models that mimic reality as closely as possible is at best naive, and at worst blatantly wrong.

The fact is that Prof. Lewin was guiding the students to think of an electromagnetic field based model (Faraday's Law), essentially invoking the spirit of Maxwell's equations, which blow any "lumped parameter model" out of the water, at least in terms of better matching the real physics.

 

[atyy]

@stevenb: MIT seems to be rewriting history, 5 minutes into http://ocw.mit.edu/courses/electric...tronics-spring-2007/video-lectures/lecture-2/, he gives the zero definition of KVL (admittedly this is what I learned too, I wasn't aware of the history of this till your earlier posts)!

 

[yungman]

So after all these pages, and many words, I think I finally realize that we all might be saying the same thing. (Well most of us, some people that didn't stick around past page 4 said some pretty weird things).

Basically, my problem with Lewin's example was he didn't account for the inductance of the wire, making it look like a node had two voltages. From an electrical engineers perspective, that is heresy, as we always attempt to make a model that mimics reality as close as possible, be it lumped element, or what. I am getting the feeling that physicist's sometimes disregard some aspects of a circuit, but I don't have a good sense on why yet.

All the talk about Lewin setting up a strange definition of KVL is making more sense to me, as KVL does not hold, IFF you claim there is no inductor lumped element.

I still contend this makes Lewin look like he doesn't know how energy is being coupled into the circuit, or is being a totally misleading jester...

All in all, I will continue to use the lumped element models for cases like this, as this is most definitely the "most intuitive" thing to do for me :).

I know what Cabraham is driving at that in case their is no wire and the whole loop is made up of resistors. Please read my post at #125. I was thinking if we can solder resistor back to back with very very short leads. so the loop is mainly consists of resistor body and see what is the measurement. You might have better access with resistors than me since I am not working and I can't access to a lab. It would be very nice if you can find some long thin resistors. but if not, you can still use those half watt carbon resistors and solder a loop. Make sure you don't use any of the wire wound resistors because they are really inductors in reasonable high frequency. You have to use either metal film or carbon resistors. Use one with about 900ohm and 6 or 8 of identical low value to make up the 100ohm. Solder tightly to a loop.

This time instead of using a scope probe, take two wires, fix one side and use an electric drill clamp on the other end and twist into a twisted pair of wires. Use the twisted pair in place of the probe and solder onto the resistors. Do the measurement and see. The reason of the twisted pair is to minimize the loop area of the prob so you don't pick up mag field. This way you get much better measurement. You can experiment waving the twisted pair around to see whether the voltage change also.

If you have two channel scope, you might have to useing two twisted pair and do measurement on two resistors at a time. You just walk through the resistors like:

1) First measurement is on R1 and R2 together and write down the voltage.
2) then move one pair to R3 and write down both reading of R2 and R3.
3) Then more the pair on R2 to R4 and write down again both reading of R3 and R4.

You repeat the measurement until you reach the last resistor and back to R1 again.

The reason for doing this is because when you solder and unsolder onto the resistor, you move the setup and even you input identical mag field, when the loop move, induction coupling change. If you do this walking through the resistor one at a time like I described, you can back calculate the voltage ratio instead of depending on the absolute reading.

When you build the resistors loop, twist the leads of the resistors together before you solder. You can leave the twisted end a little longer because it is not part of the loop and should not change the reading. This is because when you solder the leads on and off, you might undo the loop and change the characteristic when you put it back. With twisted leads, the loop will not undo when you try to solder the probe leads on.

I hope you have time to do this. This will show whether if the loop make of all resistors make a difference. I am not sure you will get the correct ratio of voltage. But I am sure it will be path independent. It this case, there will be no wire, every point is a node. We can put this whole thing behind.

I still believe Lorenze force can be modeled as a source. It is a very simple law that in the presence of mag field:

$$\vec F = q(\vec u X \vec B)$$

Can be consider as a distributed source and can be modeled as a lump source. Because this say nothing but current being moved in the presence of mag field...Nothing more. I attach a drawing.

 

[yungman]

I did enlighten you. Look at the two statements that I quoted from you. You don't see the contradiction? I can enlighten you more if you can't see it.

Lumped models are not models that mimic reality as closely as possible. The fact is that we use approximate models all the time, both in engineering and in phyisics, so your statement that electrical engineers always use models that mimic reality as closely as possible is at best naive, and at worst blatantly wrong.

The fact is that Prof. Lewin was guiding the students to think of an electromagnetic field based model (Faraday's Law), essentially invoking the spirit of Maxwell's equations, which blow any "lumped parameter model" out of the water, at least in terms of better matching the real physics.

If you like, you can use distributed model of putting infinite voltage sources spread around the whole loop and draw out the circuit. Remember now, it is the EM books that use this distributed model to model the transmission line. They model as R per unit length, C per unit length etc. Then they make the length approach zero to derive the wave equation.

You draw the distribute model of the loop out, you can see immediately that you can just move all the distributed source to one side and become a lumped source. See my drawing.

I am not sure in the resistor loop, you will get the voltage ratio of the resistors. You can argue the Ohm's law don't work, but I argue that it could be because the voltage source ins laced inside the resistor that can throw the reading off. I am not sure in this case, but it would be interesting to see if Sarumonkee is willing to do the resistors loop experiment.

 

[stevenb]

If you like, you can use distributed model of putting infinite voltage sources spread around the whole loop and draw out the circuit.

Sure, I wouldn't object to that at all. I'm also not opposed to a single lumping of EMF in some cases. It all depends on what you are trying to do.

 

[hikaru1221]

A video demonstration by MIT, which might be some help (see section 10.0.1): http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-013-electromagnetics-and-applications-fall-2005/textbook-with-video-demonstrations/

@yungman: IMHO, the reason we can group all the elements is that they are continuously distributed. However, in this case, the NET (total) emf is disrupted. Have a look at my picture. The induced emf is continuously distributed, but the extra emfs due to charge accumulation are not.
When we write the equation for a loop, since the E-field of the charge is conservative, the extra emf vanishes for a complete loop, and that's why we have such things as equivalence between KVL and FL, as stevenb and cabraham (if I'm not wrong) pointed out.

@stevenb: Hurray to you for pointing out the subtlety behind this debate.

 

[sarumonkee]

A video demonstration by MIT, which might be some help (see section 10.0.1): http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-013-electromagnetics-and-applications-fall-2005/textbook-with-video-demonstrations/

@yungman: IMHO, the reason we can group all the elements is that they are continuously distributed. However, in this case, the NET (total) emf is disrupted. Have a look at my picture. The induced emf is continuously distributed, but the extra emfs due to charge accumulation are not.
When we write the equation for a loop, since the E-field of the charge is conservative, the extra emf vanishes for a complete loop, and that's why we have such things as equivalence between KVL and FL, as stevenb and cabraham (if I'm not wrong) pointed out.

@stevenb: Hurray to you for pointing out the subtlety behind this debate.

Do the extra EMF and induced EMF equal each other in this picture?

 

[hikaru1221]

Do the extra EMF and induced EMF equal each other in this picture?

Theoretically, the part of induced emf on the wire cancels out the extra emf #1 and #2. Notice the direction. I'm on vacation and don't have access to the lab, so I'm still waiting for your experiment for verification

 

[Studiot]

Now we are all into back patting mode perhaps we should look seriously at yungman's transmission lines.

Some of these, extending perhaps across half a continent or more, are sufficiently long that the conditions at one part do not have time to affect a remote part if we take simultaneous 'readings' .

@cabraham

Thank you for noticing my comment about multiple mesh loops. This is not a problem if you take the original galvanic version of KVL for each loop.

@yungman
If you talk to the engineers at your works I expect they will be working in terms of admittances rather than resitances/impedances and using different mesh analysis methods anyway.

Oh and can anyone tell me why I now have a pink label?